Vietnamese Mathematical Olympiad

a. First, we prove by induction that $0 < x_n < \frac{3}{2n}$, $\forall n \ge 1$.
The base case $n = 1$ is trivial. For $n = 2$, we have
$$0 < x_1^2(3 - 2x_1) < 3x_1^2 < \frac{3}{4} \Rightarrow 0 < x_2 < \frac{3}{4}.$$
Assume that $0 < x_k < \frac{3}{2k}$ for some $k \ge 2$, by the AM-GM inequality,
$$0 < x_k^2(3 - 2k x_k) = \frac{1}{k^2}(k x_k)(k x_k)(3 - 2k x_k) \le \frac{1}{k^2} < \frac{3}{2(k+1)}.$$
Therefore, $0 < x_{k+1} < \frac{3}{2(k+1)}$. The inductive step is completed.
By the Squeeze theorem, we have $\lim_{n \to \infty} x_n = 0$.

b. It is clear that $x_{n+1} = 3x_n^2 - 2n x_n^3 \le 3x_n^2$ implies
$$x_{n+2} \le 3x_{n+1}^2 \le 27x_n^4 \le \frac{3^7}{16 n^4} = \frac{C}{n^4}, \forall n \ge 1.$$
For all $n > 2$, we obtain that
$$y_n = \sum_{i=1}^{n} i x_i = x_1 + 2x_2 + \sum_{i=1}^{n-2} (i+2)x_{i+2} \\ \le x_1 + 2x_2 + C \sum_{i=1}^{n-2} \frac{i+2}{i^4} \le x_1 + 2x_2 + 3C \sum_{i=1}^{n-2} \frac{1}{i^2}.$$
Hence, $(y_n)$ is bounded above. Note that $(y_n)$ is increasing, we conclude that $(y_n)$ is convergent. $\square$